Q1(a). Average and most probable distance of the 1s electron

Answer in brief

rmp = a₀ = 52.9 pm and ⟨r⟩ = (3/2)a₀ = 79.4 pm, so ⟨r⟩ = 1.5 rmp; the radial distribution curve is not symmetric and its long tail pulls the average outward.

Key points

• The radial distribution function is P(r) = 4πr²ψ₁ₛ² = (4/a₀³) r² e^−2r/a₀.
• The most probable distance is where P(r) is highest: dP/dr = 0 gives rmp = a₀ = 52.9 pm.
• The average distance is the mean of r weighted by P(r): ⟨r⟩ = (3/2)a₀ = 79.4 pm.
• On the left the curve rises steeply (r² factor); on the right it dies away slowly as e^−2r/a₀, so 67.7 % of the probability lies beyond a₀.
• ψ₁ₛ² is largest at the nucleus, but the shell volume 4πr²dr is zero there, so P(r) peaks at a₀.